CF1106C.Lunar New Year and Number Division

Lunar New Year is approaching, and Bob is struggling with his homework – a number division problem.

There are $n$ positive integers $a_1, a_2, \ldots, a_n$ on Bob's homework paper, where $n$ is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least $2$ numbers. Suppose the numbers are divided into $m$ groups, and the sum of the numbers in the $j$-th group is $s_j$. Bob's aim is to minimize the sum of the square of $s_j$, that is $$\sum_{j = 1}^{m} s_j^2.$$

Bob is puzzled by this hard problem. Could you please help him solve it?

农历新年即将来临，而鲍勃正为他的家庭作业——一道数的划分问题——而苦恼。

鲍勃的作业纸上写着 $n$ 个正整数 $a_1, a_2, \ldots, a_n$，其中 $n$ 总是偶数。鲍勃需要将这些数划分为若干组，且每组至少包含 $2$ 个数。假设这些数被划分为 $m$ 组，第 $j$ 组中所有数的和为 $s_j$。鲍勃的目标是最小化所有 $s_j$ 的平方和，即

$$\sum_{j = 1}^{m} s_j^2.$$

这道难题令鲍勃十分困惑。你能帮他解决吗？

The first line contains an even integer $n$ ($2 \leq n \leq 3 \cdot 10^5$), denoting that there are $n$ integers on Bob's homework paper.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^4$), describing the numbers you need to deal with.

第一行包含一个偶数 $n$（$2 \leq n \leq 3 \cdot 10^5$），表示鲍勃的作业纸上共有 $n$ 个整数。

第二行包含 $n$ 个整数 $a_1, a_2, \ldots, a_n$（$1 \leq a_i \leq 10^4$），描述你需要处理的数字。

A single line containing one integer, denoting the minimum of the sum of the square of $s_j$, which is $$\sum_{i = j}^{m} s_j^2,$$ where $m$ is the number of groups.

一行，包含一个整数，表示 $s_j$ 平方和的最小值，即 $$\sum_{i = j}^{m} s_j^2,$$ 其中 $m$ 为组的数量。

In the first sample, one of the optimal solutions is to divide those $4$ numbers into $2$ groups ${2, 8}, {5, 3}$. Thus the answer is $(2 + 8)^2 + (5 + 3)^2 = 164$.

In the second sample, one of the optimal solutions is to divide those $6$ numbers into $3$ groups ${1, 2}, {1, 2}, {1, 2}$. Thus the answer is $(1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27$.

在第一个样例中，一种最优解是将这 $4$ 个数划分为 $2$ 组：${2, 8}$ 和 ${5, 3}$。因此答案为 $(2 + 8)^2 + (5 + 3)^2 = 164$。

在第二个样例中，一种最优解是将这 $6$ 个数划分为 $3$ 组：${1, 2}$、${1, 2}$ 和 ${1, 2}$。因此答案为 $(1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27$。